Label the pools: 4 affirmative-only members (A), 3 negative-only members (N), and 2 either-side members (E). The panel needs exactly 2 on affirmative and exactly 2 on negative, and assigning an E member to a different side counts as a different panel. Split the count by how many E members are used.
No E member used: choose 2 from A and 2 from N, giving C(4,2) × C(3,2) = 6 × 3 = 18.
Exactly one E member used: pick which E member (2 ways), then place it. If it goes to affirmative, you still need 1 more from A and 2 from N: C(4,1) × C(3,2) = 4 × 3 = 12. If it goes to negative, you need 2 from A and 1 from N: C(4,2) × C(3,1) = 6 × 3 = 18. Each E member contributes 12 + 18 = 30 panels, and there are 2 such members, giving 2 × 30 = 60.
Both E members used: if both go to affirmative, you need 2 from N to fill negative, giving C(3,2) = 3. If both go to negative, you need 2 from A to fill affirmative, giving C(4,2) = 6. If one E member goes to each side, there are 2 ways to split them, and each split still needs 1 more from A (4 ways) and 1 more from N (3 ways), giving 2 × 4 × 3 = 24. The both-E-members total is 3 + 6 + 24 = 33.
Add the three cases: 18 + 60 + 33 = 111. The answer is D.
Choice E (150) computes the naive product C(6,2) × C(5,2), treating the pool sizes as 6 and 5 without removing panels that place the same either-side member on both sides at once. Choice B (60) keeps only the exactly-one-E-member case and drops the other two. Choice C (99) counts unordered groups of 4 people instead of side-assigned panels, missing that each group containing both E members plus one A member and one N member can be assigned to sides in 2 distinct ways. Choice A (18) omits the either-side members from the count entirely. Because every choice is a bare number, checking any of them means rebuilding all three cases yourself.